Question

The slope of the normal to the curve $x=a(\theta +sin\theta ),y=acos\theta at\theta =\frac{\pi }{4}$ is

• A.
• B. -1
• C. 1
• D.

Answer 1

The correct option is D

$x=a(\theta +sin\theta )y=acos\theta \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{-asin\theta }{a(1+cos\theta )}=\frac{-sin\theta }{a(1+cos\theta )}\text{Slope of normal at}\beta =\frac{\pi }{4}=\frac{1}{{\frac{\frac{dy}{dx}}{\theta }}_{=}\frac{\pi }{4}}=\frac{-1}{\frac{-\left(\frac{1}{\sqrt{2}}\right)}{(1+\frac{1}{\sqrt{2}})}}=(\sqrt{2}+1)$